Digits Count

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题意：问A~B之间，每个数字出现的多少次.
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解法：水题，直接计算每一位上出现的 0~9 可能的出现次数， 就是前面乘上后面的（10^k形式），然后统计即可。
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队友代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int cnt[10], left[10], right[10], digs[10];
void count(int a, int b) {
    if (a == 0) return;
    int tmp = a, len = 0, carry = 1;
    for (int i = 0; i < 10; i++) left[i] = 1, digs[i] = 0;
    right[0] = 1;
    for (int i = 1; i < 10; i++) right[i] = right[i - 1] * 10;
    while (tmp > 0) {
        left[++len] = tmp; digs[len] = digs[len - 1] + tmp % 10 * carry;
        tmp /= 10; carry *= 10;
    }
    int now = len;
    while (a > 0) { 
        if (now == len) {
            if (now > 1) {
                cnt[0] += b * (left[len - now + 2]);
                for (int i = 1; i <= a % 10; i++) {
                    cnt[i] += b * (left[len - now + 2] + 1);
                }
                for (int i = a % 10 + 1; i < 10; i++) {
                    cnt[i] += b * (left[len - now + 2]);
                }
            }
            else {
                for (int i = 1; i <= a % 10; i++) {
                    cnt[i]+=b;
                }
            }
        }
        else if (now == 1) {
            for (int i = 1; i < a % 10; i++) {
                cnt[i] += b * (right[len - now]);
            }
            cnt[a % 10] += b * (digs[len - now] + 1);
        }
        else { 
            if (a % 10 == 0) cnt[0] += b * ((left[len - now + 2] - 1) * (right[len - now]) + digs[len - now] + 1);
            else cnt[0] += b * (left[len - now + 2] ) * right[len - now];
            for (int i = 1; i < a % 10; i ++) {
                cnt[i] += b * (left[len - now + 2] + 1) * right[len - now];
            }
            if (a % 10 != 0) cnt[a % 10] += b * ((left[len - now + 2]) * right[len - now] + digs[len - now] + 1);
            for (int i = a % 10 + 1; i < 10; i++) {
                cnt[i] += b * (left[len - now + 2] ) * right[len - now]; 
            }
        }
        now--; a /= 10;  
    }
}
int main() {
    int a, b;
    while (~scanf("%d %d", &a, &b) && a) {
        memset(cnt, 0, sizeof(cnt));
        count(b, 1); count(a - 1, -1);
        for (int i = 0; i < 9; i++) printf("%d ", cnt[i]);
        printf("%d\n", cnt[9]);
    }
}